Why is the union of all proper subgroups of G equal to G?

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"Let G be a non-trivial and non-cyclic Group and let T be the union of all proper subgroups of G. Prove that G=T."

I'm a bit stuck on this problem... My first idea was that : $\forall g \in G$, we can search a subgroup $H \subseteq G$ such that $g \in H$. But I don't really know how to get this done, or even if this is a good way of starting the problem...

Note to clarify : G can be infinite and a proper subgroup cannot be equal to G. (I'm specifying this because I saw on wikipedia that some people use "proper" the same way as "non-trivial".)

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If $g \in G$ Think about how you can use the set generated by $g$, $$\langle g\rangle = \{x \in G \mid x = g^n \text{ for some }n \in \mathbb{Z} \}$$ to solve this...